Question

If ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$, then $4s(s-a)(s-b)(s-c)=$
A. ${{s}^{4}}$
B. ${{b}^{2}}{{c}^{2}}$
C. ${{c}^{2}}{{a}^{2}}$
D. ${{a}^{2}}{{b}^{2}}$

Answer 1

Hint: To solve this question, we will use the formula of the area of the right angled triangle and Heron’s formula. We will take Heron's formula and write it as the given equation, value of which we have to derive. Then by substituting the area of the right angled triangle in the derived equation we will determine the value of $4s(s-a)(s-b)(s-c)$.

Formula Used: Heron’s Formula : $Area=\sqrt{s(s-a)(s-b)(s-c)}$
Area of right-angled triangle: $\Delta =\dfrac{1}{2}\times b\times h$ where $b$ is base and $h$ is height.

Complete step by step solution: We are given that ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$ and we have to calculate the value of $4s(s-a)(s-b)(s-c)$.
As we know, Pythagoras Theorem ${{H}^{2}}={{P}^{2}}+{{B}^{2}}$ states the relation between the sides of the right angled triangle and we are given ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$, where $b$ is base, $a$ is perpendicular or height and $c$ is hypotenuse. Then it means that the given triangle is a right angled triangle.
Now,
First we will take area of right angled triangle,
$\Delta =\dfrac{1}{2}\times a\times b..........(i)$
We know that area can be also calculated with the help of Heron’s formula. So,
\[\begin{align}
  & \Delta =\sqrt{s(s-a)(s-b)(s-c)} \\
 & {{\Delta }^{2}}=s(s-a)(s-b)(s-c)
\end{align}\]
We will multiply the above equation by $4$ on both sides of the equation.
$4{{\Delta }^{2}}=4s(s-a)(s-b)(s-c)$
Now we will substitute the value of equation (i) in the above equation.
$\begin{align}
  & 4{{\left( \dfrac{1}{2}ab \right)}^{2}}=4s(s-a)(s-b)(s-c) \\
 & 4\times \dfrac{1}{4}{{a}^{2}}{{b}^{2}}=4s(s-a)(s-b)(s-c) \\
 & {{a}^{2}}{{b}^{2}}=4s(s-a)(s-b)(s-c)
\end{align}$
So we can say that the value of the equation is ${{a}^{2}}{{b}^{2}}$.

The value of $4s(s-a)(s-b)(s-c) $when ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$ is ${{a}^{2}}{{b}^{2}}$. Hence the correct option is (D).

Note: We could have also determine the value of equation $4s(s-a)(s-b)(s-c)$ by using the formula of semi perimeter of the triangle $s=\dfrac{a+b+c}{2}$.
We will substitute the value of $s$ in equation $4s(s-a)(s-b)(s-c)$ and simplify.
$\begin{align}
  & 4s(s-a)(s-b)(s-c)=4\left( \dfrac{a+b+c}{2} \right)\left( \dfrac{a+b+c}{2}-a \right)\left( \dfrac{a+b+c}{2}-b \right)\left( \dfrac{a+b+c}{2}-c \right) \\
 & =\dfrac{4}{16}(a+b+c)(b+c-a)(a+c-b)(a+b-c) \\
 & =\dfrac{1}{4}(a+b+c)(b+c-a)(a+c-b)(a+b-c)
\end{align}$
After simplifying the above equation using the expansion formula we will get the result $4s(s-a)(s-b)(s-c)={{a}^{2}}{{b}^{2}}$.